What is inverse oflaplace (10)/(2s+10) ?

The answer to inverse oflaplace (10)/(2s+10) is 5e^{-5t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (10)/(2s+10)

inverse laplace

\frac{10}{2 s + 10}

5 e^{- 5 t}

Solution steps

L^{- 1} {\frac{10}{2 s + 10}}

= L^{- 1} {5 \cdot \frac{1}{s + 5}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= 5 L^{- 1} {\frac{1}{s + 5}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{1}{s - a}} = e^{a t}

L^{- 1} {\frac{1}{s + 5}} = e^{- 5 t}

= 5 e^{- 5 t}

\int \frac{cos ( \frac{1}{x} )}{x ^{2}} d x

(lo g_{10} (n))^{2}

x y^{^{'}} - y = \frac{3}{2} x ln (x)

\int \frac{5 x - 1}{x ^{2} - 3 x - 40} d x

f (x) = \frac{x}{8 - x}, at x = 3

What is inverse oflaplace (10)/(2s+10) ?
The answer to inverse oflaplace (10)/(2s+10) is 5e^{-5t}